Friday, June 15, 2012

Mech Speed

In considering the height, I decided to reduce the length of the legs segments to 4' each instead of 5'. That will put the maximum height of the leg sections at 8'. Along with a 5' cockpit/hip assembly and a 1' foot and ankle assembly this keeps the total walking height at under 14'. Which is the requirement of the US government to drive on the roads. I still have to work out how I'll get the feet to be so short, but I'll deal with that later.

It's also clear the changes have increased the efficiency of the legs. New efficiency is 8.0255. Stride of 6.33' with a actuator displacement of 0.79'.

What does that efficiency mean to us? We want to calculate the mech speed based on the stats we now know.

Lets start with the size of the actuators we're going to need. Assume the mech weights about 2,000lbs. At worst the leg will be at a 90 deg angle and the entire weight of the mech will be at a 4x leverage so each actuator will need to handle 8,000lbs. An actuator with a 3" diameter running at 1500psi will generate 10,602lbs of force so that will do.

At 3" diameter with a 0.79' total stroke per step we get 0.29gallons displaced per step. If we have an engine generating 5gal/s (~250lbs) this will make a step time of 3.48 seconds and a speed of 1.24 MPH. With a 10gal/s (~500lbs) engine this will be a step ever 1.74 seconds and a walking speed of 2.48MPH.

Of course, I also intend to have a 10gal accumulator. It will reduce in pressure as it drains so lets say I only get 5gal out of it at >1500psi. This works out to about 17 steps or about 109' before I'll be pulling from the engine again. There's no way to know exactly how fast a sprint will be at this point because that depends on just how quickly the actuators can fill and move. It may also depend on controlling the momentum of the device as we don't want the mech falling over when it stops.